Chi-square test and F-test

CFA level I / Quantitative Methods: Application / Hypothesis Testing / Chi-square test and F-test

The variances and standard deviation are used widely in investment analysis as measures of risk. So, it is advisable for analysts to understand the hypothesis tests concerning variance.

Hypothesis test concerning the variance of a normally distributed population: A chi-square test is used to test the hypothesis whether the variance is equal to, greater than or equal to, or less than or equal to a certain value. Three hypotheses tests can be formulated.

H₀: σ² = σ²₀ versus H_a: σ² ≠ σ²₀ (two-tailed test)
H₀: σ² ≤ σ²₀ versus H_a: σ² > σ²₀ (right-tailed test)
H₀: σ² ≥ σ²₀ versus H_a: σ² < σ²₀ (left-tailed test)

The chi-square distribution is bounded below by zero, and it does not take any negative values.

test-statistic, χ² = (n-1)s²/σ²₀ with n-1 degrees of freedom

s² is the sample variance and is calculated using the normal variance formula.

The chi-square test is sensitive to violations of its assumption. If the sample is not actually random and does not come from a normally distributed population, then the inferences based on the test are likely to be faulty.

Chi-square table: The chi-square table has values for the probability in the right tail for various degrees of freedom for a different level of significance. 

If the level of significance is 0.05 and the test is a two-tailed test, then the critical values will be χ_0.025 and χ_0.975, and the null hypothesis will be rejected if the test statistic value does not fall between these two values.
 
If the level of significance is 0.05 and the test is a right-tailed test, then the critical value will be χ_0.05, and the null hypothesis will be rejected if the test statistic value is greater than the critical value.

If the level of significance is 0.05 and the test is a left-tailed test, then the critical value will be χ_0.95, and the null hypothesis will be rejected if the test statistic value is lesser than the critical value.

Example 6: Chi-square test

You want to check whether the monthly variance of an equity mutual fund is greater than 0.0050. You take a sample of 24 months. The returns are normally distributed. The sample variance is equal to 0.0062. Formulate the null and alternative hypothesis and check it for 0.05 level of significance. Solution: The chi-square test will be used, and the hypotheses are following: H0: σ2 ≤ 0.0050 versus Ha: σ2 > 0.0050 (right-tailed test) χ2 = (n-1)s2/σ20 = (24-1)0.0062/0.0050 = 28.52 The critical value for the right-tailed test for 0.05 level of significance for 23 degrees of freedom is 35.172. Since the test statistic value is lesser than the critical value, we fail to reject the null hypothesis. The variance of the monthly return of the mutual fund is statistically not greater than 0.0050 at 0.05 level of significance.

Hypothesis test concerning the equality (inequality) of two variances drawn from two normally distributed populations using two independent random samples: The F-test is used for this purpose. We can formulate three hypotheses in this case.

H₀: σ²₁ = σ²₂ versus H_a: σ²₁ ≠ σ²₂ (two-tailed test)
H₀: σ²₁ ≤ σ²₂ versus H_a: σ²₁ > σ²₂ (one-tailed test)
H₀: σ²₁ ≥ σ²₂ versus H_a: σ²₁ < σ²₂ (one-tailed test)

Test-statistic, F = s₁²/s₂²

The F-distribution is also bounded from below by zero like chi-distribution. Each F-distribution is defined by two degrees of freedom - the numerator and denominator degrees of freedom. The F-test is also sensitive to the violations of assumptions. The numerator and denominator degrees of freedom are n₁ - 1 and n₂ - 1 respectively where n₁ and n₂  are sample sizes of population 1 and population 2 respectively.

The rejection point for the null hypothesis of equality will be F_α/2, and it will be rejected if the F-statistic value is greater than the critical value. The rejection point for the null hypothesis of greater than or equal to and less than or equal to will be F_α and it will be rejected if the test statistic value is greater than the critical value.

While calculating the F-statistics, the higher of the s₁² and s₂² will be used in the numerator, and the lower value will be used in the denominator. So, its value will always be greater than 1.

If the F-statistic value is given as less than one then also we can use the F-tables by taking the reciprocal of that value and then using the tables accordingly.

Example 7: F-test

You want to test whether the volatility of returns of equity index is greater on the expiration day than other normal days for Dow Jones Industrial Average (DJIA). For that, you take the data for 30 stocks that constitute the DJIA. The sample of normal trading days has 121observations with a standard deviation of 0.748 percent. The sample of expiration days has 12 observation with a standard deviation of 0.892 percent. Formulate the null and alternative hypotheses and test the hypotheses at 0.05 level of significance. The returns on normal days and expiration days follow a normal distribution. Solution: We are comparing the variances of two normally distributed populations. So, F-test will be used. The hypotheses for the test are given below: H0: σ21 ≤ σ22 versus Ha: σ21 > σ22 where σ21 and σ22  are the variances of the returns on expiration days and normal days respectively. Test statistic, F = s12/s22 = 0.8922/0.7482 = 1.422. The F-critical value is F11,120 (numerator has 11 degrees of freedom and denominator has 120 degrees of freedom), and it is equal to 1.87 from the F-table of 0.05 level of significance. Since the test statistic is lower than the F-critical value; we fail to reject the null hypothesis. At 0.05 level of significance, the volatility on the expirations days is not statistically greater than the volatility on the normal days of trading.

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